What is the extraneous solution to these equations? $\dfrac{x^2 - 61}{x - 8} = \dfrac{-x + 11}{x - 8}$
Multiply both sides by $x - 8$ $ \dfrac{x^2 - 61}{x - 8} (x - 8) = \dfrac{-x + 11}{x - 8} (x - 8)$ $ x^2 - 61 = -x + 11$ Subtract $-x + 11$ from both sides: $ x^2 - 61 - (-x + 11) = -x + 11 - (-x + 11)$ $ x^2 - 61 + x - 11 = 0$ $ x^2 - 72 + x = 0$ Factor the expression: $ (x + 9)(x - 8) = 0$ Therefore $x = -9$ or $x = 8$ At $x = 8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 8$, it is an extraneous solution.